Question

f(x)=x^3-2x^2+3x and g(x)=3f(x+2)-1 List the transformations of f(x) to g(x) and obtain an algebraic representation of g(x)

Answer 1

First, shift right 2, then vertical stretch of 3, and finally a shift to the left 1.
\[g(x)=3(x+2)^{3} - 3(x+2)^{2} +3(3x+6) - 1\]

Answer 2

Why is the last part 3(3x+6)? Wouldn't it be 3(3x+2)? And I'm not really sure how you got the transformations?

Answer 3

it is 3x + 6 because I distributed the 3(x +2) then multiplied by the 3*f(x+2) portion.

The transformations are read from the operators on f(x) and orders of operations.
f(x+2) will shift LEFT (or negative in the x direction), the 3(fx) will stretch, and the f(x)-1 shifts DOWN (or negative in the y direction).

In review, I noticed I typed the transformations wrong the first time. Use these instead.

Answer 4

Ok. Thanks. I got the same transformations you did. But I'm still confused about the substituting parts. I thought that for every x in the f(x) function I substitute (x+2). So, without distributing, wouldn't it look like 3((x+2)^3-2(x+2)^2+3(x+2))-1? What happened with the coefficients for each term?

Answer 5

OH gah, I forgot to attach the 2 in the second, however I did distribute for the last term.
f(x)=x^3-2x^2+3x
f(x+2)=(x+2)^3-2(x+2)^2+3(x+2)
I distributed the 3 in the last form.
f(x+2)=(x+2)^3-2(x+2)^2+3x+6
Then multiply all by 3 and subtract 1.

g(x)=3(x+2)^3−6(x+2)^2+3(3x+6)−1

Answer 6

Thank you!

Answer 7

Any time :D